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2 years ago

What is the nth term of this number sequence 25, 20, 15, 10, 5

Mathematics

1 answer:

Bas_tet [7]2 years ago

6 0

Answer:

the next term you subtract 5 just as it is in the others to get the next term

Step-by-step explanation:

so5-5=0

the next term is 0

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11 ft 15 ft 13 ft what is the volumn

MakcuM [25]

Answer:

2145 ft^3

Step-by-step explanation:

7 0

2 years ago

A statistics professor plans classes so carefully that the lengths of her classes are uniformly distributed between 46.046.0 and

Gala2k [10]

Answer:

The probability of selecting a class that runs between 50.2550.25 and 51.2551.25 minutes is 0.10

Step-by-step explanation:

The Uniform Distribution, also known as Rectangular Distribution, is a type of Continuous Probability Distribution. It has a continuous random variable restricted to a finite interval and its probability function has a constant density during this interval.

The formula of probability if given by:

f(x)=

$\left \{ {{\frac{1}{b-a}; \ a \leq x \leq b } \atop {0}; \ x \ otherwise } \right.$

In this exercise a= 46.0 and b= 56.0

The probability of selecting a class that runs between 50.2550.25 and 51.2551.25 minutes is:

$\int\limits^{51.25}_{50.25} {\frac{1}{56-46} } \, dx = \int\limits^{51.25}_{50.25} {\frac{1}{10} } \, dx = \frac{1}{10} \times (51.25 - 50.25)=\frac{1}{10}=0.1$

7 0

2 years ago

If g(x)=4x-6 then g^-1(x)=? (1) -4x+6 (2)-1/4x+1/6 (3)1/4x+3/2 (4)1/4x+6

kati45 [8]

Given:

$g(x)=4x-6$

To find:

The function $g^{-1}(x)$ .

Solution:

we have,

$g(x)=4x-6$

Put g(x)=y.

$y=4x-6$

Interchange x and y.

$x=4y-6$

Isolate y.

$x+6=4y$

$\dfrac{x+6}{4}=y$

$y=\dfrac{x+6}{4}$

Substitute $y=g^{-1}(x)$ .

$g^{-1}(x)=\dfrac{x+6}{4}$

$g^{-1}(x)=\dfrac{x}{4}+\dfrac{6}{4}$

$g^{-1}(x)=\dfrac{x}{4}+\dfrac{3}{2}$

Therefore, the correct option is (3).

4 0

2 years ago

Read 2 more answers

Solve 3x2 + 4x = −5.

Alla [95]

Move the -5 over by adding 5 to both sides
3x^2+4x+5=0
must use quadratic formula

for an equation in the form
ax^2+bx+c=0
x= $\frac{-b+/- \sqrt{b^2-4ac} }{2a}$

a=3
b=4
c=5

x= $\frac{-4+/- \sqrt{4^2-4(3)(5)} }{2(3)}$
x= $\frac{-4+/- \sqrt{16-60} }{6}$
x= $\frac{-4+/- \sqrt{-44} }{6}$
x= $\frac{-4+/- 2\sqrt{-11} }{6}$
remember that√-1=i
x= $\frac{-4+/- 2i\sqrt{-11} }{6}$
x= $\frac{-2+/- i\sqrt{-11} }{3}$

x= $\frac{-2+ i\sqrt{-11} }{3}$ or x= $\frac{-2- i\sqrt{-11} }{3}$

4 0

2 years ago

What are the missing values in the following geometric sequence? __, 6, 3, __, __.

9966 [12]

The missing numbers are 12, 1.5, and 0.75 because it gets divided by 2 (or multiplied by 0.5) each time.

hope this helps.

7 0

3 years ago